Question: Let $f(x)=\dfrac{1}{x^2}$. $f'(5)=$
Solution: The strategy We can first rewrite $f(x)$ as a negative power of $x$. Then, the derivative of $f$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $f'(x)$, we can plug $x=5$ into it to find $f'(5)$. Rewriting the fraction as a negative power $f(x)=\dfrac{1}{x^2}=x^{-2}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-2}\right) \\\\ &=-2x^{-2-1} \gray{\text{The power rule}} \\\\ &=-2x^{-3} \end{aligned}$ Evaluating $f'(x)$ So we found that $f'(x)=-2x^{-3}$, which can also be written as $-\dfrac{2}{x^3}$. Now let's plug ${x=5}$ : $\begin{aligned} -\dfrac{2}{({5})^3}&=-\dfrac{2}{125} \end{aligned}$ In conclusion, $f'(5)=-\dfrac{2}{125}$.